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9th May 2016, 05:08 AM  #1 
diyAudio Member
Join Date: Aug 2009
Location: Sacramento

Single End Transfomers, The hysteresis loop and maximum current
Hello All,
Scratching my head about the way audio output transformer manufactures provide transformer specifications that call out a maximum primary current. It seems to me that selecting a transformer based on maximum current is odd. Also seems to me that the idle current selected should fall at the center of the transformer’s hysteresis loop for least distortion. I just received a pair of Transendar 10 watt, 90 ma maximum transformers. Where do you suppose that 90 ma maximum fits on the hysteresis loop? Thanks DT Last edited by DualTriode; 9th May 2016 at 05:12 AM. Reason: spelling 
9th May 2016, 10:57 AM  #2 
diyAudio Member
Join Date: May 2007

The centre of the hysteresis loop is zero DC  not much use for SE! Max current for an SE OPT is probably where the primary inductance starts falling too low. Less likely is the point where copper heating gets too much.
Remember that when AC is present the secondary current almost matches changes in the primary current, so flux stays almost the same. 
9th May 2016, 11:09 AM  #3 
diyAudio Member
Join Date: Jun 2005
Location: San Diego, CA

It's called the "minor loop" where the AC flux circulates...on top of the DC offset..
You have the SUM of the DC and AC flux density..... Lets suppose you want to limit the total flux to 10k Gauss.. The DC flux density can be at 5K Gauss and the AC minor loop will swing 5k Gauss..meaning up to 10K and down to 0 origin.... If you want to stay off the origin since it is nonlinear in this region.. you can DC bias at 6k Gauss and then have the AC flux swing 4k Gauss up to 10k Gauss and down to 2k Gauss.. The more turns you add the greater the DC flux will increase, but the AC flux goes down, so you have a smaller minor loop... 
9th May 2016, 04:41 PM  #4  
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Join Date: Oct 2010
Location: Traslasierra

Quote:
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9th May 2016, 05:12 PM  #5 
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Join Date: Nov 2003
Location: Brighton UK


9th May 2016, 05:46 PM  #6  
diyAudio Member
Join Date: Oct 2010
Location: Traslasierra

Quote:
AC magnetic field Bac(max) = (Uac x 10⁸) / (√2 π fo S Np) DC magnetic field Bdc(max) = [4 π μ Np i(DC)] / (9 l) Where i(DC) = Primary DC current [A] l = Magnetic circuit length [cm] Np=Number of primary turns S=Core area μ=Magnetic permeability including the air gap i(DC) is provided by the manufacturer, it is i(DC) not 2i(DC)
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9th May 2016, 07:01 PM  #7 
diyAudio Member RIP
Join Date: Nov 2003
Location: Brighton UK

Hi,
Well I assumed wrong. Fair enough, A toss up between the two. Saturation current of the core is easy to measure, so you specify that, or half that for single ended use. Rethinking it makes far more sense to specify SE standing current. rgds, sreten. 
9th May 2016, 08:12 PM  #8 
diyAudio Member
Join Date: Aug 2009
Location: Sacramento

So we have it. We have theory and formulas.
We have standing idle DC current. On top of the DC idle current we have AC signal current. For the greatest symmetrical AC current swing what will the standing DC idle current be for this Transendar 10 watt 90ma maximum current transformer? Or put another way 30ma standing DC current may be too low. 100ma may be too high? I am guessing that the stated 90ma maximum may be pushing up against the too high end for optimal symmetrical AC current swing. I am looking for the DC current that allows for the greatest symmetrical AC current swing. DT 
9th May 2016, 09:40 PM  #9 
diyAudio Member RIP
Join Date: Nov 2003
Location: Brighton UK

Hi,
Then you need to know the saturation current and simply halve it. Which I suspect is the given current rating. Your overcomplicating. rgds, sreten. 
9th May 2016, 10:16 PM  #10  
diyAudio Member
Join Date: Oct 2010
Location: Traslasierra

Quote:
2. There is no defined "saturation current", transformer core saturates due to magnetic field B, Bdc(max) or Idc(max) is a designer's choice. It follows that you do not need to halve nothing. Quote:
Transformer manufacturer say that if you put more than 90 mA DC, in the best case you will not obtain 10W, in the worst case you will saturate (more probable) and even magnetize permanently your transformer core (less probable). As the effect of DC current is move the hysteresis loop on the plane BH, the best result is obtained close to nominal 90 mA, then idealy (well designed trafo) the hysteresis loop will be in the I cuadrant and you can avoid zero crossing.
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